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The standard equation of a circle is \((x−h)^2+(y−k)^2=r^2\), where the point (h,k) is the center of the circle and r represents the length of the radius. If a particular circle can be represented by the equation \(x^2+y^2−8x+2y=−1\), what is the radius of that circle?

Re: The standard equation of a circle is (x−h)^2+(y−k)^2=r^2, where the [#permalink]

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29 Nov 2015, 10:23

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x^2 + y^2 - 8x + 2y = -1 To bring the above equation into its standard form, complete the squares by adding 16 and 1 on both the sides x^2 - 8x + 16 + y^2 + 2y + 1 = -1 + 16 + 1 (x - 4)^2 + (y + 1)^2 = 16 --> This is in the form of (x-h)^2 + (y-k)^2 = r^2 r = 4

Re: The standard equation of a circle is (x−h)^2+(y−k)^2=r^2, where the [#permalink]

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03 Feb 2016, 17:14

Hi Bunuel,

This is my first time participating in gmatclub. Firs of all thank you for always posting stuff that really helps and for always giving a good solution to math problems.

I've been struggling trying to find out an easy procedure for this question. Can you help?

Re: The standard equation of a circle is (x−h)^2+(y−k)^2=r^2, where the [#permalink]

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03 Feb 2016, 18:08

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betancurj wrote:

Hi Bunuel,

This is my first time participating in gmatclub. Firs of all thank you for always posting stuff that really helps and for always giving a good solution to math problems.

I've been struggling trying to find out an easy procedure for this question. Can you help?

Thanks again

Juan

Let me try to answer.

You are given that the standard equation of a circle ---> (x−h)^2+(y−k)^2=r^2 where, r=radius of the circle.

Now, you are given another circle with the equation ---> x^2 + y^2 - 8x + 2y = -1 , when you compare this to the standard equation of the circle, you should realize that it would be straightforward to calculate the radius of the circle if you can express x^2 + y^2 - 8x + 2y = -1 in the form of (x−h)^2+(y−k)^2=r^2

In order to make complete squares similar to (x-h)^2 or (y-k)^2, you need to remember that (x+a)^2=x^2+a^2+2ax and (x-a)^2=x^2+a^2-2ax

With these relations in mind, lets go back to x^2 + y^2 - 8x + 2y = -1 with a focus on converting x^2 + y^2 - 8x + 2y to 2 complete squares (1 with x and 1 with y).

x^2 + y^2 - 8x + 2y ---> (x^2-8x+16-16)+(y^2+2y+1-1) = (x-4)^2 -16 + (y+1)^2-1 ---> x^2 + y^2 - 8x + 2y = -1 can thus be written as :

The standard equation of a circle is (x−h)^2+(y−k)^2=r^2, where the [#permalink]

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03 Feb 2016, 20:28

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betancurj wrote:

Hi Bunuel,

This is my first time participating in gmatclub. Firs of all thank you for always posting stuff that really helps and for always giving a good solution to math problems.

I've been struggling trying to find out an easy procedure for this question. Can you help?

Thanks again

Juan

Hi Juan,

The trick here is to convert everything in the standard form of equation of a circle, which is: \((x−h)^2+(y−k)^2=r^2\)

We are given the equation of a circle in some other form (expanded form of the standard equation)

\(x^2+y^2−8x+2y=−1\) On grouping terms,

\(x^2−8x+y^2+2y=−1\) From here on, we need to make two perfect squares in x and y

The best way to understand which perfect square to make is by dissecting the coefficient of x and y 8 = 2*1*4, 2 = 2*1*1

We are writing the coefficients in this form, because the coefficient of x in \({(x+a)}^2 = x^2 + 2ax + a^2\) (Eq i) is 2*1*a \(x^2−2*x*4 + y^2+2*1*y = -1\)

Now we need the constant terms to make these perfect squares. The constant term in Eq (i) is a

Hence we can write \(x^2−2*1*4x + y^2+2*1*y = -1\) as \(x^2−2*x*4 + 16 + y^2+2*1*y + 1 = -1 + 16 + 1\) (adding constants on both sides of the equation) \({(x-4)}^2 + {(y+1)}^2 = 16\) \({(x-4)}^2 + {(y+1)}^2 = 4^2\)

Re: The standard equation of a circle is (x−h)^2+(y−k)^2=r^2, where the [#permalink]

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15 Jun 2017, 23:35

Bunuel wrote:

The standard equation of a circle is \((x−h)^2+(y−k)^2=r^2\), where the point (h,k) is the center of the circle and r represents the length of the radius. If a particular circle can be represented by the equation \(x^2+y^2−8x+2y=−1\), what is the radius of that circle?

Then don't H and K have to be 4 and -1? Unless I'm making some assumption here that wouldn't necessarily apply to some other example- please let me know

Re: The standard equation of a circle is (x−h)^2+(y−k)^2=r^2, where the [#permalink]

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15 Jun 2017, 23:44

OptimusPrepJanielle wrote:

betancurj wrote:

Hi Bunuel,

This is my first time participating in gmatclub. Firs of all thank you for always posting stuff that really helps and for always giving a good solution to math problems.

I've been struggling trying to find out an easy procedure for this question. Can you help?

Thanks again

Juan

Hi Juan,

The trick here is to convert everything in the standard form of equation of a circle, which is: \((x−h)^2+(y−k)^2=r^2\)

We are given the equation of a circle in some other form (expanded form of the standard equation)

\(x^2+y^2−8x+2y=−1\) On grouping terms,

\(x^2−8x+y^2+2y=−1\) From here on, we need to make two perfect squares in x and y

The best way to understand which perfect square to make is by dissecting the coefficient of x and y 8 = 2*1*4, 2 = 2*1*1

We are writing the coefficients in this form, because the coefficient of x in \({(x+a)}^2 = x^2 + 2ax + a^2\) (Eq i) is 2*1*a \(x^2−2*x*4 + y^2+2*1*y = -1\)

Now we need the constant terms to make these perfect squares. The constant term in Eq (i) is a

Hence we can write \(x^2−2*1*4x + y^2+2*1*y = -1\) as \(x^2−2*x*4 + 16 + y^2+2*1*y + 1 = -1 + 16 + 1\) (adding constants on both sides of the equation) \({(x-4)}^2 + {(y+1)}^2 = 16\) \({(x-4)}^2 + {(y+1)}^2 = 4^2\)

Hence the radius = 4. Option D

Yes but don't forget to explain you derived 16 and 1 by taking half of the coefficients 8 and 1 and then squaring them. For example, if we have an algebraic expression that cannot be factored such as x-10x-18=0 - then we complete the square. How?

x-10x __ = 18 (add 18 to both sides or really just put 18 on the other side) x-10x +25 = 18 + 25 (remember how to get the coefficients? In this equation a= 1 b= 10; take (b/2)^2 and add to both sides (x-5)(x-5) = 43 x-5 = + or minus 43

The standard equation of a circle is \((x−h)^2+(y−k)^2=r^2\), where the point (h,k) is the center of the circle and r represents the length of the radius. If a particular circle can be represented by the equation \(x^2+y^2−8x+2y=−1\), what is the radius of that circle?

What should first jump out is that the equation for this particular circle does not match the standard equation of a circle. So you'll need to do some algebra to get to that point.

Your first step should be to factor the x and y terms so that they fit the standard form. Since each term needs to be factored into a squared parenthetical, you should use the common quadratic form \((a−b)^2=a^2−2ab+b^2\) as your guide. Look at the coefficients −8x and +2y, recognizing that they need to fill the −2ab role in that common quadratic equation. This means that you'll factor the x and y terms into:

\((x−4)^2=x^2−8x+16\) and \((y−(−1)^2)=y^2+2y+1\)

So now you know that the left-hand side of the standard equation of a circle produces the combination of \(x^2−8x+16+y^2+2y+1\), which equals \(x^2−8x+y^2+2y+17\). So now you know two equations to be true:

\(x^2−8x+y^2+2y+17=r^2\)

and

\(x^2+y^2−8x+2y=−1\)

That second equation can be quickly made to look like the first by adding 1 to both sides:

\(x^2+y^2−8x+2y+1=0\)

Which means that the difference between the two equations is that the left-hand side is 16 apart and the right-hand side is r^2 apart. So to balance, add 16 to the left and r^2 to the right, meaning that r^2=16. This means that the radius is 4, making answer choice D correct.
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